# Can water in solid state (ice) be at such a low temperature that in contact with liquid water at 0, freezes it and leaves it at a lower than zero C°?

I recently had a discussion with a bio engineer and we are both pretty convinced of different outcomes for the same situation:

Say you have a glass half full of liquid water at 0°, could you get a piece of ice off no more volume than the water in the glass that is at such a low temperature that it takes the water to freeze stabilize the whole glass content way lower than 0°c (say -20c°)?

My bet is that could be if the piece of ice is cold enough (-200°c? maybe not that cold?) you could. I believe I'm right from what I remember of thermodynamics in university, but I didn't get my title and he did.

if it's possible and the ice/ water volume (not mass) is the same and you can depreciate the glass and its temperature which temp would be a rough estimate that would get you that (let's say stabilize at -20°c)?

if that's not possible and the liquid water could never get lower than 0° in a solid state could you explain why?

The heat of fusion of water is 80 cal/gm and the heat capacity of ice is about 0.5 cal/gm-C. If you had 100 gm of liquid water at 0 C and 100 gm of ice, it would take 8000 cal. to freeze the water. The temperature change of the ice to do this would be 8000/(100)(0.5)=160 C. So the ice would have to be at about -160 C initially to bring this about, and the final temperature would be 0C. This neglects the heat needed to cool the glass.

• I did a quick search and according to one source the specific heat of ice is lower at lower temperatures. The lowest entry in the table is for -100 C. and it gives a value of 1.4 $kJ{kg}^{-1}{K}^{-1}$. That is significantly lower than the value of about 2 for ice close to 0 C. So it may be the case that even ice at close to 0 K. does not have enough capacity to absorb enough heat from the water to freeze it. It's a close call, by the looks of it Commented Mar 21, 2021 at 14:17
• That's interesting. Commented Mar 21, 2021 at 16:46
• @Cleonis That info isn't easy to find! But I finally hit paydirt: The heat capacity of water ice in interstellar or interplanetary conditions by L. M. Shulman (2004). He gives an equation derived by combining the Debye-Sommerfeld equation with experimental data from 1936 and 1960: $$C_p = 7.73×10^{-3}T \left(1-e^{-1.263×10^{-3}T^2}\right) × \left(1+e^{-3\sqrt T} × 8.47×10^{-3}T^6 + 2.0825×10^{-7}T^4 e^{-4.97×10^{-2}T}\right) \; Jg^{-1}K^{-1}$$ Commented Mar 21, 2021 at 17:32
• @Cleonis Done! It turns out that even ice near 0 K doesn't have enough capacity to freeze an equal mass of water. Commented Mar 21, 2021 at 21:52
• Sure, or a little over 11% more, by mass. Commented Mar 21, 2021 at 22:01

Somewhat surprisingly, your Bioengineer friend is correct. A given volume of ice, even if the ice's temperature is almost 0 K (absolute zero), does not have a sufficiently high enthalpy to absorb enough heat from an equal volume of water at 0° C to freeze it all into ice.

The problem is that ice has a lower specific heat capacity than liquid water. From that Wikipedia article:

Liquid water has one of the highest specific heats among common substances, about 4182 J/(K kg) at 20 °C; but that of ice just below 0 °C is only 2093 J/(K kg).

At lower temperatures, the specific heat capacity of ice is even lower, approaching zero as the temperature approaches 0 K.

In order to freeze water at 0° C, we need to remove energy equal to its enthalpy of fusion (aka latent heat of fusion), which is $$335.55 \,\mathrm{kJ/kg}$$. However, ice at 0° C (~273.15 K) has enthalpy just under $$300.31 \,\mathrm{kJ/kg}$$. So to completely freeze 1 kg of water you need a little over 1.11 kg of ice near absolute zero, and the final temperature will be close to 0° C, not counting losses due to heat from the environment or the container.

The volume of 1 kg of water at 0° C is close to 1 litre. Ice at that temperature has a lower density than water, and the mass of 1 litre of 0° C ice is close to 0.9168 kg. The density of ice (at standard atmospheric pressure) increases slightly as the temperature drops. Eg, at -100° C, its density is around 0.9257 kg / L. Unfortunately, I can't find density data for ice near 0 K, but I doubt that it gets up to 1.11 kg / L. Ice has various crystal structures that dominate at different temperature and pressure regimes, please see Wikipedia for details.

It's not easy to find information on the specific heat capacity of ice at low temperatures. The best source I found is The heat capacity of water ice in interstellar or interplanetary conditions by L. M. Shulman (2004). In that paper, the author gives a formula for the $$C_p$$ (the specific heat capacity) of ice, derived from the Debye-Sommerfeld equation combined with experimental data from 1936 and 1960:

$$C_p = 7.73×10^{-3}T \left(1-e^{-1.263×10^{-3}T^2}\right) × \left(1+e^{-3\sqrt T} × 8.47×10^{-3}T^6 + 2.0825×10^{-7}T^4 e^{-4.97×10^{-2}T}\right) \; Jg^{-1}K^{-1}$$

Note that this equation uses mass units of grams, not kilograms.

To obtain the enthalpy, we need to integrate the specific heat capacity: $$\int_0^{T'} C_p\,dT$$

For our purposes, we need $$T'=273.15$$

I calculated the enthalpy given above by doing numerical integration using the free open-source mathematics software system, SageMath. I also used SageMath to create some graphs using that equation.

### Enthalpy

Here's my Sage / Python script, which can generate plots in SVG format or as PNG images.

Assuming the entire system is thermally insulated, then you could get the entire frozen glass below freezing in this way. If once everything freezes the section of the original ice block is still below freezing then temperature of the entire system will continue to drop until the entire system is at the same temperature (thermal equilibrium) below freezing. This is just the heat equation (heat diffusion) at work.

There is no principle saying water cannot get below freezing.

It is of course possible, in general, to use ice to freeze water. But the water-to-ice ratio is too big or if water is too hot would the attempt fail. Your specific case is borderline, if ice and water have the same mass (similar volume) or same volume (similar mass) it might be hard but by increasing the amount of ice you might make it.

Let's explore it in details in a simple model of a substance which has constant heat capacity - the real case is discussed in between. See also PM 2Ring's answer for more details.

When put in contact, the ice will absorb a quantity of heat $$Q$$ from the water and increase its temperature and the water will donate $$Q$$ to the ice, thus lowering its own temperature. So the question is simply, whether the ice can take that quantity $$Q$$ of heat without melting.

We assume the system is isolated, and there is no loss of heat. The only heat transfer is between ice and water. We also assme everything happens slow enough. Finally, as we like to ave $$0C$$ as a reference temperature we will work with Celsius degrees rather the Kelvin.

We will first do the math and the comment the results at the end, with some numbers

The model

If we have a mass $$M_i$$ of ice at temperature $$T_i$$ and $$M_w$$ of water at temperature $$T_w$$ and we put them into contact, they will start exchanging heat.

Water will cool down and reach a temperature $$T_f$$. The heat it will have donated in doing that will be $$Q_w=M_wc_w(T_f-T_w)$$ where $$c_w$$ is the specific heat of water. if $$T_f<0$$ i.e. if water has turned into ice, it will donate an extra amount $$-\lambda M_w$$ of heat, where $$\lambda$$ is the specific latent heat i.e. the amount of heat each unit of mass gives while changing phase.

So if we start with water at $$T_w\ge 0$$ and end up with water at $$T_f<0$$, the heat given by the water is $$Q_w=M_wc_w(T_f-T_w)-\lambda M_w<0$$.

I am going to use the same value $$c_w$$ for the whole transition, but if one wants to be extra precise, one can also assume that ice has a different heat capacity than water and then we split that into a part from $$T_w$$ to $$T=0$$ with $$c_w$$ and one from $$T=0$$ to $$T_f$$ with a different value $$c_i$$ which is the heat capacity of ice. So: $$Q_w=Q_w=-M_wc_wT_w+M_w c_i T_f-\lambda M_w$$

Even more specifically, because the heat capacity is a function of the temperature $$c(T)$$ you would in principle need to sum all contributions at different temperature so the best formula for the heat exchange would be $$Q=-\lambda M_w +\int_{T_w}^{T_i} M_wc(T)dT$$ Also, because $$c(T)$$ is very low for ice near absolute $$0$$, cold ice will not resist absorption very efficiently compared to "warm ice" and that should be taken into account for precise results. However from now on, for the sake of simplicity, we ignore this detail, the final numerical result will be "wrong" but the concept will stay the same.

[Apparently (see answer by PM 2Ring) by putting in the exact formula for $$c(T)$$ you find that if you start with the same amount of ice and water you will not make it. So you need more ice.]

Anyways, water donates $$Q$$ and this heat is absorbed by the ice. Ice was starting at temperature $$T_i$$ and ends up at $$T_f$$ becase at the end water and ice have the same temperature. The maximum heat that ice can absorb without melting would be if $$T_f<0$$ (even if by a very little bit) and it would be given by

$$Q_{MAX}=M_i c_w (T_f-T_i)$$ (again using $$c_w$$ for ice too! Otherwise here use $$c_i$$).

So the condition for the system to exchange heat without the ice melting but rather with the water freezing would be if the total heat the water donates is smaller than the maximum heat the ice can absorb i.e. if

$$|Q_w|<|Q_{MAX}|$$ so $$|M_wc_w(T_f-T_w)-\lambda M_w| < |M_i c_w (T_f-T_i)|$$

Because we know that $$T_i we can rewrite it eliminating the absolute value as

$$\lambda M_w+M_wc_w(T_w-T_f) < M_i c_w (T_f-T_i)$$

So by putting $$T_f=0$$ we get the minimum temperature your ice needs to be at when you put it in contact with water so that the whole system, after having exchanged heat is still at $$0$$ degrees. Any extra fraction of degree less of your ice would lead to a completely forzen system at the end. That limit temperature is

$$T_m=-{\lambda M_w\over c_w M_i} -{M_w\over M_i}T_w$$

So given a mass of ice $$M_i$$ and of water $$M_w$$ starting at $$T_w$$ then the whole thing will freeze if $$T_i.

While the actual value for $$c_w$$ should be adjusted by computing the integral above, this formula sums up all the contributions to our problem: the latent heat $$\lambda$$ (the minimun heat you need to freeze water to ice at $$T=0C$$, the mass of ice and water (the more water you have, the more ice you need) and the specific heat of water (and ice) telling you how much resistance to heat exchange a given mass of water (ice) has before changing its temperature. The higher this value is for water (and the lower it is for ice) the more ice you need (because ice resists very little to heat exchanges before melting compared to water which can absorb a lot of heat before lowering its temperature significantly)

First of all, let me reiterate that because of the dependency of $$c(T)$$ on temperature our results are wrong. But I am keeping it to show you what the general issues in this problem are.

If we assume $$M_w=M_i$$ (same mass) and, as you said, $$T_w=0$$, we get

$$T_m=-{\lambda \over c_w}$$ and that is, using $$\lambda=334 J / g$$ and $$c_w\approx 3 J/g C$$ (an average of the specific heat of water and of ice, although considering we are more on the ice side it would be maybe closer to 1-2.. - $$C$$ is for Celsius ) then you get that ice must have a temperature of at least

$$T_m=-{\lambda \over c_w}=-{334 J /g \over 3 J /g }C=-111 C$$

(I neglected a 0.33 degree for simplicity).

Notice that, from the general formula above, you also get that if your water starts at $$T_w>0$$ then for each extra degree of water temperature you need one extra less degree of initial ice temperature. This limit value of $$-111C$$ is the temperature a mass of ice needs to have to absorb, without melting, the latent heat of an equivalent mass of water turning to ice. It's sort of a limit temperature which guarantess the minimum reservoir of heat the ice needs to have in order to absorb that of water, so for equal masses the final formula is

$$T_m=-{\lambda \over c_w}-T_w=-(T_w + 111 C)$$

However, if you have $$M_i=2M_w$$ (two times more ice than water), again starting at $$T_w=0$$ you get that you can use "hotter" ice as the heat now is stored in mass rather than in temperature

$$T_m=-{\lambda \over 2 c_w} = -55.5 C$$

and more in general, for a mass of ice $$n$$ times the mass of water ($$M_i=nM_w$$) starting at $$T_w$$

$$T_m=-{\lambda \over n c_w} = -{1\over n}( T_w + 111 C)$$

As before, if the initial ice if even colder than that, then the final equilibrium temperature will be less than zero, so not only can you freeze water, you can also cool it below zero with ice if it is enough or at least cold enough.

Also, you can freeze air vapor if you want with the same procedure, but you are going to need more, colder ice.

However, because there is a limit to the temperature (-273 C = 0K) if you start with water which is too hot or if you $$M_w\gg M_i$$ (there is too much water in comparison to ice) then it might not be possible to freeze the water unless you add more and more ice. Again, if your mass of ice is the same of the mass of water, you can not freeze the water if $$T_w > 163 C$$ (at which point it would be vapor, but whatever). [If our guess for $$c_w\approx 2$$ is wrong, as it apparently is and then $$c_w$$ is actually closer to 0 than that, for this reason, you might not be able to use ice to freeze water unless you use a bigger mass of ice.

• I did a quick search for the specific heats of ice and water respectively, and according to one source the specific heat of ice (temperature close to 0 C.) is about 2 $J{kg}^{-1}{K}^{-1}$ and that of water (close to 0 C.) is about 4 $J{kg}^{-1}{K}^{-1}.$ So it seems worthwhile to treat those two specific heats as separate. Also the value of the specific heat can be quite different in different temperature ranges. Commented Mar 21, 2021 at 13:56
• Yes but the final formula would not be so "elegant" (: the real formula for the exchanged heat would be $$Q=\int_{T_i}^{T_f} M_w c(T) dT-\lambda M_w$$ .. I think it won't change much to the general principle, only the value of $T_m$ would change a bit... Commented Mar 21, 2021 at 13:58
• From en.wikipedia.org/wiki/Adsorption "Adsorption is a surface phenomenon, while absorption involves the whole volume of the material" Commented Mar 21, 2021 at 14:18
• You really do need to take into account the lower heat capacity of ice, and the fact that its heat capacity reduces even further as you approach absolute zero. Commented Mar 21, 2021 at 21:57
• I am surprised, I have to say: you are right, apparently. The best thing would be to merge the answers and give an "effective" value for the mean heat capacity using your formula... I have edited my answer taking this into account and clarifying the assumptions, I still think the answer shows "the logic" nice enough, then it is a mater of putting in the real numbers of course. Commented Mar 23, 2021 at 9:23